Question

Evaluate the definite integral: $∫^1_{-1}(x+1)dx$

Answer 1

The correct answer is: 2
Let $I=∫^1_{-1}(x+1)dx$
$∫(x+1)dx=\frac{x^2}{2}+x=F(x)$
By second fundamental theorem of calculus,we obtain
$I=F(1)-F(-1)$
$=(\frac{1}{2}+1)-(\frac{1}{2}-1)$
$=\frac{1}{2}+1-\frac{1}{2}+1$
$=2$